AS-Level Maths: Core 1 for Edexcel C1.2 Algebra and functions 2 This icon indicates the slide contains activities created in Flash. These activities are not editable. For more detailed instructions, see the Getting Started presentation. 1 of 50 © Boardworks Ltd 2005 Quadratic expressions Quadratic expressions Contents Factorizing quadratics Completing the square Solving quadratic equations The discriminant Graphs of quadratic functions Examination-style questions 2 of 50 © Boardworks Ltd 2005 Quadratic expressions A quadratic expression is an expression in which the highest power of the variable is 2. For example: x2 – 2 w2 + 3w + 1 4 – 5g2 t2 2 The general form of a quadratic expression in x is: ax2 + bx + c (where a ≠ 0) x is a variable. a is the coefficient of x2. b is the coefficient of x. c is a constant term. 3 of 50 © Boardworks Ltd 2005 Factorizing quadratics Quadratic expressions Contents Factorizing quadratics Completing the square Solving quadratic equations The discriminant Graphs of quadratic functions Examination-style questions 4 of 50 © Boardworks Ltd 2005 Factorizing quadratic expressions Factorizing an expression is the inverse of expanding it. Expanding or multiplying out x2 + 3x + 2 (x + 1)(x + 2) Factorizing When we expand an expression we multiply out the brackets. When we factorize an expression we write it with brackets. 5 of 50 © Boardworks Ltd 2005 Factorizing quadratic expressions No constant term Quadratic expressions of the form ax2 + bx can always be factorized by taking out the common factor x. For example: 3x2 – 5x = x(3x – 5) The difference between two squares When a quadratic has no term in x and the other two terms can be written as the difference between two squares, we can use the identity a2 – b2 = (a + b)(a – b) to factorize it. For example: 9x2 – 49 = (3x + 7)(3x – 7) 6 of 50 © Boardworks Ltd 2005 Factorizing quadratic expressions Quadratic expressions with a = 1 Quadratic expressions of the form x2 + bx + c can be factorized if they can be written using brackets as (x + d)(x + e) where d and e are integers. If we expand (x + d)(x + e), we have (x + d)(x + e) = x2 + dx + ex + de = x2 + (d + e)x + de 7 of 50 © Boardworks Ltd 2005 Factorizing quadratic expressions The general form Quadratic expressions of the general form ax2 + bx + c can be factorized if they can be written using brackets as (dx + e)(fx + g) where d, e, f and g are integers. If we expand (dx + e)(fx + g), we have (dx + e)(fx + g)= dfx2 + dgx + efx + eg = dfx2 + (dg + ef)x + eg 8 of 50 © Boardworks Ltd 2005 Completing the square Quadratic expressions Contents Factorizing quadratics Completing the square Solving quadratic equations The discriminant Graphs of quadratic functions Examination-style questions 9 of 50 © Boardworks Ltd 2005 Perfect squares Some quadratic expressions can be written as perfect squares. For example: x2 + 2x + 1 = (x + 1)2 x2 – 2x + 1 = (x – 1)2 x2 + 4x + 4 = (x + 2)2 x2 – 4x + 4 = (x – 2)2 x2 + 6x + 9 = (x + 3)2 x2 – 6x + 9 = (x – 3)2 In general: x2 + 2ax + a2 = (x + a)2 or x2 – 2ax + a2 = (x – a)2 How could the quadratic expression x2 + 8x be made into a perfect square? We could add 16 to it. 10 of 50 © Boardworks Ltd 2005 Completing the square Adding 16 to the expression x2 + 8x to make it into a perfect square is called completing the square. We can write x2 + 8x = x2 + 8x + 16 – 16 If we add 16 we then have to subtract 16 so that both sides are still equal. By writing x2 + 8x + 16 we have completed the square and so we can write this as x2 + 8x = (x + 4)2 – 16 In general: 2 b b x bx x 2 2 2 2 11 of 50 © Boardworks Ltd 2005 Completing the square Complete the square for x2 – 10x. Compare this expression to (x – 5)2 = x2 – 10x + 25 x2 – 10x = x2 – 10x + 25 – 25 = (x – 5)2 – 25 Complete the square for x2 + 3x. Compare this expression to ( x + 32 ) 2 = x 2 + 3 x + 2 2 x + 3x = x + 3x + 9 4 = ( x + 32 ) 9 4 2 12 of 50 9 4 9 4 © Boardworks Ltd 2005 Completing the square How can we complete the square for x2 – 8x + 7? Look at the coefficient of x. This is –8 so compare the expression to (x – 4)2 = x2 – 8x + 16. x2 – 8x + 7 = x2 – 8x + 16 – 16 + 7 = (x – 4)2 – 9 In general: 2 2 b b x bx c x c 2 2 2 13 of 50 © Boardworks Ltd 2005 Completing the square Complete the square for x2 + 12x – 5. Compare this expression to (x + 6)2 = x2 + 12x + 36 x2 + 12x – 5 = x2 + 12x + 36 – 36 – 5 = (x + 6)2 – 41 Complete the square for x2 – 5x + 7. Compare this expression to ( x 52 ) 2 = x 2 5 x + x 5x + 7 = x 5x + 2 2 = ( x + 52 ) 2 14 of 50 25 4 25 4 25 4 7 3 4 © Boardworks Ltd 2005 Completing the square When the coefficient of x2 is not 1, quadratic equations in the form ax2 + bx + c can be rewritten in the form a(x + p)2 + q by completing the square. Complete the square for 2x2 + 8x + 3. Take out the coefficient of x2 as a factor from the terms in x: 2x2 + 8x + 3 = 2(x2 + 4x) + 3 By completing the square, x2 + 4x = (x + 2)2 – 4 so 2x2 + 8x + 3 = 2((x + 2)2 – 4) + 3 = 2(x + 2)2 – 8 + 3 = 2(x + 2)2 – 5 15 of 50 © Boardworks Ltd 2005 Completing the square Complete the square for 5 + 6x – 3x2. Take out the coefficient of x2 as a factor from the terms in x: 5 + 6x – 3x2 = 5 – 3(–2x + x2) = 5 – 3(x2 – 2x) By completing the square, x2 – 2x = (x – 1)2 – 1 so 5 + 6x – 3x2 = 5 – 3((x – 1)2 – 1) = 5 – 3(x – 1)2 + 3 = 8 – 3(x – 1)2 16 of 50 © Boardworks Ltd 2005 Complete the square 17 of 50 © Boardworks Ltd 2005 Solving quadratic equations Contents Quadratic expressions Factorizing quadratics Completing the square Solving quadratic equations The discriminant Graphs of quadratic functions Examination-style questions 18 of 50 © Boardworks Ltd 2005 Quadratic equations The general form of a quadratic equation in x is: ax2 + bx + c = 0 (where a ≠ 0) Quadratic equations can be solved by: factorization completing the square, or using the quadratic formula. The solutions to a quadratic equation are called the roots of the equation. A quadratic equation may have: two real distinct roots one repeated root, or no real roots. 19 of 50 © Boardworks Ltd 2005 The roots of a quadratic equation If we sketch the graph of a quadratic function y = ax2 + bx + c the roots of the equation coincide with the points where the function cuts the x-axis. As can be seen here, this can happen twice, once or not at all. 20 of 50 © Boardworks Ltd 2005 Solving quadratic equations by factorization Solve the equation 5x2 = 3x Start by rearranging the equation so that the terms are on the left-hand side: Don’t divide 5x2 – 3x = 0 through by x! Factorizing the left-hand side gives us x(5x – 3) = 0 So x=0 or 5x – 3 = 0 5x = 3 x= 21 of 50 3 5 © Boardworks Ltd 2005 Solving quadratic equations by factorization Solve the equation x2 – 5x = –4 by factorization. Start by rearranging the equation so that the terms are on the left-hand side. x2 – 5x + 4 = 0 We need to find two integers that add together to make –5 and multiply together to make 4. Because 4 is positive and –5 is negative, both the integers must be negative. These are –1 and –4. Factorizing the left-hand side gives us (x – 1)(x – 4) = 0 x–1=0 x–4=0 or x=1 x=4 22 of 50 © Boardworks Ltd 2005 Solving quadratics by completing the square Quadratic equations that cannot be solved by factorization can be solved by completing the square. For example, the quadratic equation x2 – 4x – 3 = 0 can be solved by completing the square as follows: (x – 2)2 – 7 = 0 (x – 2)2 = 7 x–2= 7 x=2+ 7 x = 4.65 23 of 50 or x=2– 7 x = –0.646 (to 3 s.f.) © Boardworks Ltd 2005 Solving quadratics by completing the square Solve the equation 2x2 – 4x + 1 = 0 by completing the square. Write the answer to 3 significant figures. Start by completing the square for 2x2 – 4x + 1: 2x2 – 4x + 1 = 2(x2 – 2x) + 1 = 2((x – 1)2 – 1) + 1 = 2(x – 1)2 – 2 + 1 = 2(x – 1)2 – 1 24 of 50 © Boardworks Ltd 2005 Solving quadratics by completing the square Now solving the equation 2x2 – 4x + 1 = 0: 2(x – 1)2 – 1 = 0 2(x – 1)2 = 1 ( x 1) = 2 1 2 x 1= x = 1+ x = 1.71 25 of 50 1 2 or 1 2 x = 1 1 2 x = 0.293 (to 3 s.f.) © Boardworks Ltd 2005 Using the quadratic equation formula Any quadratic equation of the form ax2 + bx + c = 0 can be solved by substituting the values of a, b and c into the formula x= b ± b 4 ac 2 2a This formula can be derived by completing the square on the general form of the quadratic equation. 26 of 50 © Boardworks Ltd 2005 Using the quadratic formula Use the quadratic formula to solve 2x2 + 5x – 1 = 0. 2x2 + 5x – 1 = 0 x= x= x= x= 5 + 4 x = 0.186 27 of 50 b ± b 4 ac 2 2a 5 ± 5 (4 × 2 × 1) 2 2×2 5 ± 33 25 + 8 4 or x= 5 33 4 x = –2.69 (to 3 s.f.) © Boardworks Ltd 2005 Using the quadratic formula Use the quadratic formula to solve 9x2 – 12x + 4 = 0. 9x2 – 12x + 4 = 0 x= x= x= x= b ± b 4 ac 2 2a ( 1 2 ) ± (12) (4 × 9 × 4 ) 2 2×9 12 ± 144 144 18 12 ± 0 18 There is one repeated root: x = 28 of 50 2 3 © Boardworks Ltd 2005 Equations that reduce to a quadratic form Some equations, although not quadratic, can be written in quadratic form by using a substitution. For example: Solve the equation t4 – 5t2 + 6 = 0. This is an example of a quartic equation in t. Let’s substitute x for t2: x2 – 5x + 6 = 0 This gives us a quadratic equation that can be solved by factorization: (x – 2)(x – 3) = 0 x=2 or t2 = 2 So t = 2 or 29 of 50 x=3 t2 = 3 t= 3 © Boardworks Ltd 2005 The discriminant Contents Quadratic expressions Factorizing quadratics Completing the square Solving quadratic equations The discriminant Graphs of quadratic functions Examination-style questions 30 of 50 © Boardworks Ltd 2005 The discriminant By solving quadratic equations using the formula x b b 4 ac 2 2a we can see that we can use the expression under the square root sign, b2 – 4ac, to decide how many roots there are. When b2 – 4ac > 0, there are two real distinct roots. When b2 – 4ac = 0, there is one repeated root: x b . 2a When b2 – 4ac < 0, there are no real roots. Also, when b2 – 4ac is a perfect square, the roots of the equation will be rational and the quadratic will factorize. b2 – 4ac is called the discriminant of ax2 + bx + c 31 of 50 © Boardworks Ltd 2005 The discriminant We can demonstrate each of these possibilities graphically. 32 of 50 © Boardworks Ltd 2005 Graphs of quadratic functions Contents Quadratic expressions Factorizing quadratics Completing the square Solving quadratic equations The discriminant Graphs of quadratic functions Examination-style questions 33 of 50 © Boardworks Ltd 2005 Plotting graphs of quadratic functions A quadratic function in x can be written in the form: y = ax2 + bx + c (where a ≠ 0) We can plot the graph of a quadratic function using a table of values. For example: Plot the graph of y = x2 – 4x + 2 for –1 < x < 5. x –1 0 1 2 3 4 5 x2 1 0 1 4 9 16 25 – 4x +4 +0 –4 –8 – 12 – 16 – 20 +2 y = x2 – 4x + 2 +2 7 +2 2 +2 –1 +2 –2 +2 –1 +2 2 +2 7 34 of 50 © Boardworks Ltd 2005 Plotting graphs of quadratic functions x –1 0 1 2 3 4 5 y = x2 – 4x + 2 7 2 –1 –2 –1 2 7 The points given in the table are plotted … … and the points are then joined together with a smooth curve. The shape of this curve is called a parabola. It is characteristic of a quadratic function. 35 of 50 y 6 5 4 3 2 1 –1 0 –1 1 2 3 4 5 x © Boardworks Ltd 2005 Parabolas Parabolas have a vertical axis of symmetry … …and a turning point called the vertex. When the coefficient of x2 is positive the vertex is a minimum point and the graph is -shaped. When the coefficient of x2 is negative the vertex is a maximum point and the graph is -shaped. 36 of 50 © Boardworks Ltd 2005 Exploring graphs of the form y = ax2 + bx + c 37 of 50 © Boardworks Ltd 2005 Sketching graphs of quadratic functions When a quadratic function factorizes we can use its factorized form to find where it crosses the x-axis. For example: Sketch the graph of the function y = x2 – 2x – 3. The function crosses the x-axis when y = 0. x2 – 2x – 3 = 0 (x + 1)(x – 3) = 0 x+1=0 x = –1 or x–3=0 x=3 The function crosses the x-axis at the points (–1, 0) and (3, 0). 38 of 50 © Boardworks Ltd 2005 Sketching graphs of quadratic functions By putting x = 0 in y = 2x2 – 5x – 3 we can also find where the function crosses the y-axis. y = 2(0)2 – 5(0) – 3 y=–3 So the function crosses the y-axis at the point (0, –3). In general: The quadratic function y = ax2 + bx + c will cross the y-axis at the point (0, c). We now know that the function y = x2 – 2x – 3 passes through the points (–1, 0), (3, 0) and (0, –3) and so we can place these points on our sketch. 39 of 50 © Boardworks Ltd 2005 Sketching graphs of quadratic functions We can also use the fact that a parabola is symmetrical to find the coordinates of the vertex. y (3, 0) (–1, 0) 0 x The x coordinate of the vertex is half-way between –1 and 3. (0, –3) (1, –4) When x = 1, x= 1+ 3 =1 2 y = (1)2 – 2(1) – 3 y = –4 So the coordinates of the vertex are (1, –4). We can now sketch the graph. 40 of 50 © Boardworks Ltd 2005 Sketching graphs of quadratic functions In general: When a quadratic function is written in the form y = a(x – α)(x – β), it will cut the x-axis at the points (α, 0) and (β, 0). α and β are the roots of the quadratic function. For example, write the quadratic function y = 3x2 + 4x – 4 in the form y = a(x – α)(x – β) and hence find the roots of the function. This function can be factorized as follows, y = (3x – 2)(x + 2) It can be written in the form y = a(x – p)(x – q) as y = 3( x 32 ) ( x 2 ) Therefore, the roots are 41 of 50 2 3 and 2 . © Boardworks Ltd 2005 Exploring graphs of the form y = a(x – α)(x – β) 42 of 50 © Boardworks Ltd 2005 Sketching graphs by completing the square When a function does not factorize we can write it in completed square form to find the coordinates of the vertex. For example: Sketch the graph of y = x2 + 4x – 1 by writing it in completed square form. x2 + 4x – 1 = (x + 2)2 – 5 The least value that (x + 2)2 can have is 0 because the square of a number cannot be negative. (x + 2)2 ≥ 0 Therefore (x + 2)2 – 5 ≥ – 5 The minimum value of the function y = x2 + 4x – 1 is therefore y = –5. 43 of 50 © Boardworks Ltd 2005 Sketching graphs by completing the square When y = –5, we have, (x + 2)2 – 5 = –5 (x + 2)2 = 0 x = –2 The coordinates of the vertex are therefore (–2, –5). The equation of the axis of x = –2 y symmetry is x = –2. Also, when x = 0 we have y = x2 + 4x – 1 y = –1 0 x So the curve cuts the y-axis at (–1, 0) the point (–1, 0). Using symmetry we can now sketch the graph. (–2, –5) 44 of 50 © Boardworks Ltd 2005 Sketching graphs by completing the square In general, when the quadratic function y = ax2 + bx + c is written in completed square form as a(x + p)2 + q The coordinates of the vertex will be (–p, q). The axis of symmetry will have the equation x = –p. Also: If a > 0 (–p, q) will be the minimum point. If a < 0 (–p, q) will be the maximum point. Plotting the y-intercept, (0, c) will allow the curve to be sketched using symmetry. 45 of 50 © Boardworks Ltd 2005 Exploring graphs of the form y = a(x + p)2 + q 46 of 50 © Boardworks Ltd 2005 Examination-style questions Contents Quadratic expressions Factorizing quadratics Completing the square Solving quadratic equations The discriminant Graphs of quadratic functions Examination-style questions 47 of 50 © Boardworks Ltd 2005 Examination-style question a) Write 2x2 – 8x + 7 in the form a(x + b)2 + c. b) Write down the minimum value of f(x) = 2x2 – 8x + 7 and state the minimum value of x where this occurs. c) Solve the equation 2x2 – 8x + 7 = 0 leaving your answer in surd form. d) Sketch the graph of y = 2x2 – 8x + 7. 2x2 – 8x + 7 = 2(x2 – 4x) + 7 a) = 2((x – 2)2 – 4) + 7 = 2(x – 2)2 – 8 + 7 = 2(x – 2)2 – 1 48 of 50 © Boardworks Ltd 2005 Examination-style question f(x) can be written as f(x) = 2(x – 2)2 – 1 b) From this we can see that the minimum value of f(x) is –1. This occurs when x = 2. 2x2 – 8x + 7 = 0 c) 2(x – 2)2 – 1 = 0 2(x – 2)2 = 1 (x – 2)2 = 1 2 x–2=± 1 2 x=2± x=2 49 of 50 1 2 or x=2+ 1 2 1 2 © Boardworks Ltd 2005 Examination-style question d) When y = 0, x = 2 – 1 2 or x=2+ 1 2 When x = 0, y = 7 So the graph cuts the coordinate axes at (2 + (2 – 21 , 0) and (0, 7). 1 2 , 0), The parabola has a minimum at the point (2, –1). y 7 –1 50 of 50 2– 1 2 2+ 1 2 x © Boardworks Ltd 2005

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# C1.2 Algebra and functions 2